Coordinate Geometry: Straight Lines

Competitive Mathematics · SSC CGL

5 Part33 worked examples161 practice questions

In the previous chapter we worked with points. Now we write a whole straight line as one equation. The equation is a rule: every point $(x, y)$ of the line satisfies it and no point outside the line does. For example the rule $2 x - y = 1$ fits $(1, 1)$ , $(2, 3)$ and $(3, 5)$ , and they are in one line. With this rule the meeting of two lines, the angle between lines and the distance from a point all come out.

◭ Diagram

The line

2 x - y = 1

: the coordinates of every point on it make the rule true (for

(3, 5)

6 - 5 = 1

Question	Method
How steep is the line?	Slope
What is the equation of the line?	Forms: $y = m x + c$ , point-slope, two-point, intercept
What area does it enclose with the axes?	The intercept triangle
Where do two lines meet?	Intersection
Parallel, perpendicular, the angle between?	Comparing slopes
How far is a point from a line?	Perpendicular distance formula

RememberThe formulas of the previous chapter (distance, midpoint, section, area) are used here too — keep them at hand. In every line question write two things clearly: the equation of the line and the points on it.

Slope and equation

Steepness ·

y = m x + c

· intercepts

The slope of a line, the forms of its equation and the triangle it makes with the axes.

To pin down a line you need two things: how steep it is (slope) and where it passes. That is how the equation is made.

The slope (gradient)

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = tan θ

The steepness of a line: from two points, from the angle and from the equation.

The idea

The steepness of a staircase is how far you climb for each step forward. A line is the same: divide how far you went up (or down) between two points by how far you went sideways. This quotient is the slope $m$ . On a straight line it is the same for every pair of points, which is why the line is straight. The slope also tells the angle the line makes with the positive X-axis: $m = tan θ$ , with $θ$ measured anticlockwise.

◭ Diagram

From

A (1, 1)

B (5, 4)

: run 4 (green), rise 3 (rose), so the slope is

\frac{3}{4}

; and

tan θ

of the angle

θ

is the same.

∑ Formula

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = tan θ

Slope of the line through

(x_{1}, y_{1})

and

(x_{2}, y_{2})

$m$: the slope
$x_{1}, y_{1}$: coordinates of the first point
$x_{2}, y_{2}$: coordinates of the second point
$θ$: angle from the positive X-axis, anticlockwise

∑ Formula

a x + b y = c \Rightarrow m = - \frac{a}{b}

Slope from the equation

$a, b$: coefficients of $x$ and $y$
$c$: the constant term

◭ Diagram

Four cases:

m > 0

(climbing, green),

m < 0

(falling, rose),

m = 0

(flat, blue) and a vertical line (violet) whose slope is undefined because the run is zero.

Angle $θ$	$0^{\circ}$	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$	$9 0^{\circ}$	$12 0^{\circ}$	$13 5^{\circ}$	$15 0^{\circ}$
Slope $m$	$0$	$\frac{1}{3}$	$1$	$3$	undefined	$- 3$	$- 1$	$- \frac{1}{3}$

Spot itRecognise it: “slope”, “inclination”, “angle with the X-axis”, “slope of

a x + b y = c

”, “line through two points”. If three points give equal slopes they are collinear. Trap: a vertical line (

x = k

) has no slope, and a horizontal line (

y = k

) has slope

0

Steps

Write the two points, or change the equation to $y = m x + c$ .
Take both differences in the same order: $y_{2} - y_{1}$ and $x_{2} - x_{1}$ .
Divide; if the $x$ difference is zero the line is vertical.
For the angle match $m = tan θ$ with the table; for a negative slope use $18 0^{\circ} -$ the acute angle.

Solved examples

Example 1 · Basic

Find the slope and the angle of inclination of the line through

P (- 2, 5)

and

Q (4, - 1)

Solution, step by step

1The rise: $y_{2} - y_{1} = - 1 - 5 = - 6$ (it goes 6 down).
2The run: $x_{2} - x_{1} = 4 - (- 2) = 6$ .
3 $m = \frac{- 6}{6} = - 1$ .
4 $tan θ = - 1$ is negative, so $θ$ is obtuse. $tan 4 5^{\circ} = 1$ , so $θ = 18 0^{\circ} - 4 5^{\circ} = 13 5^{\circ}$ .

Answer $m = - 1$ , $θ = 13 5^{\circ}$

A negative slope always means an angle over

9 0^{\circ}

Example 2 · Standard

The line through

A (2, 3)

and

B (k, 7)

has slope

\frac{4}{3}

. Find

k

, and also

y

on the same line at

x = 8

Solution, step by step

1Slope formula: $\frac{7 - 3}{k - 2} = \frac{4}{3}$ , i.e. $\frac{4}{k - 2} = \frac{4}{3}$ .
2Cross-multiply: $4 \times 3 = 4 (k - 2)$ , so $k - 2 = 3$ and $k = 5$ .
3On the line the slope is $\frac{4}{3}$ everywhere. From $A (2, 3)$ to $x = 8$ the run is $6$ , so the rise is $\frac{4}{3} \times 6 = 8$ .
4 $y = 3 + 8 = 11$ . Check: the slope between $(8, 11)$ and $A$ is $\frac{11 - 3}{8 - 2} = \frac{8}{6} = \frac{4}{3}$ ✓.

Answer $k = 5$ , and $y = 11$

A slope of

\frac{4}{3}

says “every 3 forward, 4 up” — reach any point on the line with it.

Example 3 · Advanced

Find the angle the line through

A (- 1, 2)

and

B (2, 2 + 33)

makes with the X-axis. Then give the inclination of a second line through

A

9 0^{\circ}

to the first.

Solution, step by step

1The run is $2 - (- 1) = 3$ and the rise is $(2 + 33) - 2 = 33$ .
2 $m = \frac{3 3}{3} = 3$ . The table has $tan 6 0^{\circ} = 3$ , so $θ = 6 0^{\circ}$ .
3The second line is turned $9 0^{\circ}$ from the first, so its inclination is $6 0^{\circ} + 9 0^{\circ} = 15 0^{\circ}$ (anticlockwise from the positive X-axis).
4Check: $tan 15 0^{\circ} = - \frac{1}{3}$ and the first slope is $3$ ; the product is $3 \times (- \frac{1}{3}) = - 1$ — the signature of perpendicular lines (seen later).

Answer $6 0^{\circ}$ ; the second line $15 0^{\circ}$

When

3

appears in a slope, think

6 0^{\circ}

12 0^{\circ}

Try it yourself

BasicTry it yourself first

Find the slope of the line through

(1, 3)

and

(5, 11)

BasicTry it yourself first

Find the slope of the line through

(- 4, 6)

and

(2, - 3)

BasicTry it yourself first

A line has slope

1

. What is its angle of inclination

θ

(measured anticlockwise from the positive X-axis)?

StandardTry it yourself first

A line has slope

- 1

. What is its angle of inclination

θ

(measured anticlockwise from the positive X-axis)?

StandardTry it yourself first

A line makes an angle of

12 0^{\circ}

with the positive X-axis. Find its slope.

StandardTry it yourself first

The line through

(3, - 2)

and

(k, 6)

has slope

2

. Find

k

StandardTry it yourself first

Find the slope of the line

5 x + 4 y = 20

StandardTry it yourself first

A road runs straight between

(- 6, - 5)

and

(3, - 2)

(each square is 1 m). Find its slope.

AdvancedTry it yourself first

A line has slope

- \frac{1}{3}

. What is its angle of inclination

θ

(measured anticlockwise from the positive X-axis)?

AdvancedTry it yourself first

Find the slope of the line

7 x - 3 y = 21

Forms of the equation of a line

y = m x + c

· point-slope · two-point · intercept

Building the equation from what is given (slope, points, intercepts) and reading slope and intercepts from an equation.

The idea

The equation of a line is the rule linking the $x$ and $y$ of each of its points. Because the slope is the same everywhere on the line, the slope between any point $(x, y)$ of the line and one known point $(x_{1}, y_{1})$ must equal $m$ : $\frac{y - y _{1}}{x - x _{1}} = m$ . This is the point-slope form, and all the other forms come from it. Which form to use depends on what the question gives.

◭ Diagram

The line

y = \frac{2}{3} x + 1

: slope

\frac{2}{3}

(up 2 over 3 across) and Y-intercept

c = 1

(violet).

Given	Form	Equation
slope $m$ and Y-intercept $c$	slope-intercept	$y = m x + c$
slope $m$ and a point $(x_{1}, y_{1})$	point-slope	$y - y_{1} = m (x - x_{1})$
two points	two-point	$y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$
cuts $a$ , $b$ on the axes	intercept	$\frac{x}{a} + \frac{y}{b} = 1$
general form	general	$a x + b y + c = 0$
parallel to an axis		$x = k$ (vertical), $y = k$ (horizontal)

Here $m$ = slope, $c$ = the Y-intercept, $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ = points on the line, $a$ and $b$ = the cuts on the X- and Y-axes.

◭ Diagram

Intercept form: the line cuts the X-axis at

a = 4

and the Y-axis at

b = - 3

, so

\frac{x}{4} + \frac{y}{- 3} = 1

, i.e.

3 x - 4 y = 12

∑ Formula

a x + b y = c \Rightarrow m = - \frac{a}{b}, X -cut = \frac{c}{a}, Y -cut = \frac{c}{b}

Reading everything from the general form

$a, b$: coefficients of $x$ and $y$
$c$: the constant on the right
$m$: the slope

Spot itRecognise it: “find the equation”, “line through …”, “equal intercepts”, “sum of intercepts”. Every time put the given point back into the equation as a check. Use the intercept form when the cuts on the axes are given; a line through the origin has no intercept form (

c = 0

Steps

See what is given and choose the form from the table above.
With two points find the slope first, then use point-slope.
Clear fractions by multiplying and write the result as $a x + b y = c$ .
Check by putting the given points in.

Solved examples

Example 4 · Basic

Write the equation of the line with slope

- \frac{3}{4}

and Y-intercept

5

in the form

a x + b y = c

Solution, step by step

1Slope-intercept form: $y = m x + c$ with $m = - \frac{3}{4}$ , $c = 5$ : $y = - \frac{3}{4} x + 5$ .
2Multiply by 4 to clear the denominator: $4 y = - 3 x + 20$ .
3Bring $x$ to the left: $3 x + 4 y = 20$ .
4Check: Y-cut: at $x = 0$ , $4 y = 20$ , $y = 5$ ✓; slope $- \frac{3}{4}$ ✓.

Answer $3 x + 4 y = 20$

y = m x + c

c

is directly the Y-cut.

Example 5 · Standard

Find the equation of the line through

(- 2, 3)

and

(4, - 5)

and both of its intercepts.

Solution, step by step

1Slope: $m = \frac{- 5 - 3}{4 - ( - 2 )} = \frac{- 8}{6} = - \frac{4}{3}$ .
2Point-slope from the first point: $y - 3 = - \frac{4}{3} (x + 2)$ .
3Multiply by 3: $3 y - 9 = - 4 x - 8$ , i.e. $4 x + 3 y = 1$ .
4Check with the second point: $4 (4) + 3 (- 5) = 16 - 15 = 1$ ✓.
5Intercepts: at $y = 0$ , $x = \frac{1}{4}$ ; at $x = 0$ , $y = \frac{1}{3}$ .

Answer $4 x + 3 y = 1$ ; X-cut $\frac{1}{4}$ , Y-cut $\frac{1}{3}$

Checking with the second point also catches mistakes made with the first.

Example 6 · Advanced

Find the lines through

(4, 3)

whose intercepts on the two axes add up to 14.

Solution, step by step

1Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ with the condition $a + b = 14$ , so $b = 14 - a$ .
2 $(4, 3)$ is on the line: $\frac{4}{a} + \frac{3}{14 - a} = 1$ . Clearing denominators: $4 (14 - a) + 3 a = a (14 - a)$ .
3Expand: $56 - a = 14 a - a^{2}$ , i.e. $a^{2} - 15 a + 56 = 0$ , which factorises as $(a - 7) (a - 8) = 0$ . So $a = 7$ or $a = 8$ .
4For $a = 7$ , $b = 7$ : $\frac{x}{7} + \frac{y}{7} = 1$ , i.e. $x + y = 7$ . For $a = 8$ , $b = 6$ : $\frac{x}{8} + \frac{y}{6} = 1$ , i.e. $3 x + 4 y = 24$ .
5Check: $4 + 3 = 7$ ✓ and $3 (4) + 4 (3) = 24$ ✓.

Answer $x + y = 7$ and $3 x + 4 y = 24$

For “equal intercepts” take

a = b

, giving

x + y = a

(or the line may pass through the origin).

Try it yourself

BasicTry it yourself first

Find the equation of the line through

(2, 5)

with slope

3

StandardTry it yourself first

Find the equation of the line through

(- 3, 4)

with slope

- \frac{2}{3}

BasicTry it yourself first

Find the equation of the line through

(1, 4)

and

(3, - 2)

StandardTry it yourself first

Find the equation of the line through

(- 4, - 1)

and

(2, 5)

StandardTry it yourself first

A line cuts the X-axis at

5

and the Y-axis at

3

. Find its equation.

StandardTry it yourself first

A line cuts the X-axis at

- 6

and the Y-axis at

4

. Find its equation.

StandardTry it yourself first

Find the Y-intercept of the line

6 x + 5 y = 30

StandardTry it yourself first

Find the X-intercept of the line

3 x - 4 y = 24

AdvancedTry it yourself first

Find the slope of the line

5 x - 2 y = 10

AdvancedTry it yourself first

The point

(6, k)

lies on the line

7 x + 3 y = 21

. Find

k

AdvancedTry it yourself first

Find the equation of the line through

(- 5, 7)

and

(3, - 1)

The triangle a line makes with the axes

Area

= \frac{c ^{2}}{2∣ ab ∣}

The intercepts of $a x + b y = c$ , the triangle with the axes, and the trapezium between two parallel lines.

The idea

The line $a x + b y = c$ cuts the axes at $\frac{c}{a}$ on the X-axis and $\frac{c}{b}$ on the Y-axis. The axes are perpendicular, so the line and the axes enclose a right triangle whose legs are these two cuts. Its area is therefore $\frac{1}{2} \times \frac{c}{a} \times \frac{c}{b}$ . This is the quickest shortcut in the chapter: straight from the equation, $\frac{c ^{2}}{2∣ ab ∣}$ , with no drawing.

◭ Diagram

The line

3 x + 5 y = 30

: cuts

\frac{30}{3} = 10

and

\frac{30}{5} = 6

, so the area is

\frac{1}{2} \times 10 \times 6 = 30

; shortcut

\frac{3 0 ^{2}}{2 \cdot 15} = 30

∑ Formula

Area = \frac{c ^{2}}{2∣ a b ∣}

Area of the triangle enclosed by the line

a x + b y = c

and the two axes

$a, b$: coefficients of $x$ and $y$
$c$: the constant on the right ( $c \neq = 0$ )
$∣ ∣$: absolute value

In the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ the area is simply $\frac{1}{2} ∣ a b ∣$ (here $a, b$ are the cuts). The trapezium of two parallel lines $a x + b y = c_{1}$ and $a x + b y = c_{2}$ with the axes is the bigger triangle minus the smaller: $\frac{∣ c _{2}^{2} - c _{1}^{2} ∣}{2∣ ab ∣}$ . The hypotenuse (the part of the line between the axes) is $a_{0}^{2} + b_{0}^{2}$ , where $a_{0}, b_{0}$ are the cuts.

◭ Diagram

The parallel lines

2 x + 3 y = 12

(green) and

2 x + 3 y = 24

(blue): triangles 12 and 48; the trapezium between them is

48 - 12 = 36

Spot itRecognise it: “area enclosed by the line and the axes”, “both coordinate axes and the line”, “product of intercepts”, “trapezium formed by two lines and the axes”. Care: if

c = 0

the line passes through the origin and no triangle forms.

Steps

Write the line as $a x + b y = c$ with $c$ on the right.
Find the area from $\frac{c ^{2}}{2∣ ab ∣}$ (or take the cuts $\frac{c}{a}$ , $\frac{c}{b}$ and use $\frac{1}{2} \times$ their product).
With two parallel lines find both triangles and subtract.
If the area is given, solve backwards for $c^{2}$ ; remember the $\pm$ .

Solved examples

Example 7 · Basic

Find the area of the triangle formed by the line

4 x + 5 y = 40

and the two coordinate axes.

Solution, step by step

1X-cut: putting $y = 0$ gives $4 x = 40$ , $x = 10$ . Y-cut: putting $x = 0$ gives $5 y = 40$ , $y = 8$ .
2The axes are perpendicular, so the cuts are the legs of a right triangle.
3Area $= \frac{1}{2} \times 10 \times 8 = 40$ sq. units.
4Shortcut check: $\frac{c ^{2}}{2∣ ab ∣} = \frac{1600}{2 \times 20} = 40$ ✓.

Answer 40 sq. units

Writing

4 x + 5 y = 40

\frac{x}{10} + \frac{y}{8} = 1

shows the cuts directly.

Example 8 · Standard

Find the area of the trapezium enclosed by the lines

5 x + 2 y = 20

5 x + 2 y = 40

and the two axes.

Solution, step by step

1Both lines have the same $a = 5$ , $b = 2$ , so they are parallel and make two triangles with the axes.
2Smaller triangle ( $c = 20$ ): $\frac{2 0 ^{2}}{2 \times 10} = \frac{400}{20} = 20$ .
3Bigger triangle ( $c = 40$ ): $\frac{4 0 ^{2}}{20} = \frac{1600}{20} = 80$ .
4The part between them is $80 - 20 = 60$ sq. units, which is the trapezium.

Answer 60 sq. units

The smaller triangle sits inside the bigger one; so subtract, do not add.

Example 9 · Advanced

Find the line through

(2, 3)

that makes a triangle of area 12 sq. units with the axes in the first quadrant.

Solution, step by step

1Take the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ ( $a, b > 0$ ). The area $\frac{1}{2} ab = 12$ , so $ab = 24$ .
2 $(2, 3)$ is on the line: $\frac{2}{a} + \frac{3}{b} = 1$ . Multiplying by $ab$ gives $2 b + 3 a = ab = 24$ .
3Put $b = \frac{24}{a}$ : $\frac{48}{a} + 3 a = 24$ . Multiplying by $a$ : $48 + 3 a^{2} = 24 a$ , i.e. $a^{2} - 8 a + 16 = 0$ , i.e. $(a - 4)^{2} = 0$ .
4 $a = 4$ and $b = \frac{24}{4} = 6$ . The line is $\frac{x}{4} + \frac{y}{6} = 1$ , i.e. $3 x + 2 y = 12$ .
5Check: $3 (2) + 2 (3) = 12$ ✓; area $\frac{1 2 ^{2}}{2 \times 6} = 12$ ✓.

Answer $3 x + 2 y = 12$ (the only such line)

Getting a perfect square shows that 12 is the smallest possible area for lines through this point.

Try it yourself

BasicTry it yourself first

Find the area (sq. units) of the triangle enclosed by the line

3 x + 2 y = 12

and the two coordinate axes.

BasicTry it yourself first

Find the area (sq. units) of the triangle enclosed by the line

5 x + 4 y = 20

and the two coordinate axes.

StandardTry it yourself first

Find the area (sq. units) of the triangle enclosed by the line

2 x - 3 y = 12

and the two coordinate axes.

StandardTry it yourself first

Find the area of the trapezium enclosed by the lines

3 x + 4 y = 12

3 x + 4 y = 24

and the two axes.

StandardTry it yourself first

Find the perimeter of the triangle enclosed by the line

\frac{x}{6} + \frac{y}{8} = 1

and the two axes.

StandardTry it yourself first

The line

2 x + 3 y = c

and the two axes enclose a triangle of area 12 sq. units (

c > 0

). Find

c

StandardTry it yourself first

Find the area (sq. units) of the triangle enclosed by the line

7 x + 2 y = 14

and the two coordinate axes.

AdvancedTry it yourself first

Find the area of the trapezium enclosed by the lines

4 x + 3 y = 24

4 x + 3 y = 36

and the two axes.

AdvancedTry it yourself first

Find the perimeter of the triangle enclosed by the line

\frac{x}{- 5} + \frac{y}{12} = 1

and the two axes.

Two lines, and a line with a point

Meeting · parallel · perpendicular · distance · image

How two lines relate (meeting, parallel, perpendicular, angle) and how a point relates to a line (distance, foot, image).

With slope and equation in hand we now look at relations between two things: line and line, and point and line.

Where two lines meet

Solve · parallel or same · concurrent · triangle

The point where two lines meet, the conditions for meeting, concurrency of three lines and the triangle they make.

The idea

The common point of two lines is the point whose $(x, y)$ makes both equations true together. So we find it by solving both equations simultaneously. There are three possibilities: the lines meet at one point (one solution), they are parallel (no solution), or they are the same line (infinitely many solutions). Which one it is shows from the ratios of their $a$ , $b$ , $c$ before solving.

◭ Diagram

The lines

3 x + y = 11

(blue) and

x - y = 1

(green) meet at

P (3, 2)

9 + 2 = 11

✓ and

3 - 2 = 1

✓. The common point is the one that fits both equations.

Case	Condition (lines $a_{1} x + b_{1} y = c_{1}$ , $a_{2} x + b_{2} y = c_{2}$ )	Result
meet at one point	$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$	one solution
parallel and distinct	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$	no solution
the same line	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$	infinitely many solutions

∑ Formula

x = \frac{c _{1} b _{2} - c _{2} b _{1}}{a _{1} b _{2} - a _{2} b _{1}}, y = \frac{a _{1} c _{2} - a _{2} c _{1}}{a _{1} b _{2} - a _{2} b _{1}}

Meeting point of

a_{1} x + b_{1} y = c_{1}

and

a_{2} x + b_{2} y = c_{2}

$a_{i}, b_{i}$: coefficients of $x$ and $y$ in the $i$ -th line
$c_{i}$: the constant of the $i$ -th line
$a_{1} b_{2} - a_{2} b_{1}$: the determinant (zero means parallel)

Three lines are concurrent (through one point) if the meeting point of the first two also satisfies the third. The triangle of three lines: find the meeting point of each pair; these are the three vertices; then use the area formula of the previous chapter. With two lines and an axis, the cuts of the two lines on the axis give the base and the distance of their meeting point from the axis is the height.

◭ Diagram

The three lines

y = x + 2

x + y = 10

and

y = 1

meet in pairs at

(4, 6)

(- 1, 1)

(9, 1)

. Base 10, height 5, area 25.

Spot itRecognise it: “point of intersection”, “where do they meet”, “crossing”, “concurrent”, “through one point”, “enclosed by three lines”, “no solution / infinitely many”. After solving, check in both equations.

Steps

Write both equations as $a x + b y = c$ and look at the ratios $\frac{a _{1}}{a _{2}}$ , $\frac{b _{1}}{b _{2}}$ (are they parallel?).
Remove one variable (multiply and subtract) and find the other.
Check by putting the values in both equations.
With three lines find each pair’s meeting point and then the area.

Solved examples

Example 10 · Basic

Find the point where the lines

2 x + 5 y = 1

and

4 x - 3 y = 15

meet.

Solution, step by step

1Compare ratios: $\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ and $\frac{b _{1}}{b _{2}} = \frac{5}{- 3}$ — not equal, so the lines meet at one point.
2To remove $x$ multiply the first by 2: $4 x + 10 y = 2$ . Subtract the second: $13 y = - 13$ .
3 $y = - 1$ . Put it in the first equation: $2 x - 5 = 1$ , so $x = 3$ .
4Check: in the second, $4 (3) - 3 (- 1) = 12 + 3 = 15$ ✓.

Answer $(3, - 1)$

Simplify the equations before eliminating.

Example 11 · Standard

For what value of

k

do the lines

k x + 3 y = 9

and

4 x + 6 y = 15

have no common point? Can they be the same line for any

k

Solution, step by step

1No common point = parallel and distinct: $\frac{k}{4} = \frac{3}{6} \neq = \frac{9}{15}$ .
2From $\frac{k}{4} = \frac{1}{2}$ , $k = 2$ .
3Check the third ratio: $\frac{9}{15} = \frac{3}{5}$ , different from $\frac{1}{2}$ ✓. So at $k = 2$ the lines are parallel and distinct, with no solution.
4To be the same line all three ratios must be equal: $\frac{1}{2} = \frac{3}{5}$ never holds, so for no $k$ are they one line.

Answer $k = 2$ ; never the same line

For

k \neq = 2

there is exactly one meeting point.

Example 12 · Advanced

Find the vertices and the area of the triangle enclosed by the lines

x - y = - 2

x + y = 10

and

y = 1

Solution, step by step

1First two lines: adding gives $2 x = 8$ , $x = 4$ , and $y = 6$ . Vertex $(4, 6)$ .
2First and $y = 1$ : $x - 1 = - 2$ , so $x = - 1$ . Vertex $(- 1, 1)$ .
3Second and $y = 1$ : $x + 1 = 10$ , so $x = 9$ . Vertex $(9, 1)$ .
4The two vertices $(- 1, 1)$ and $(9, 1)$ have the same ordinate, so this horizontal side is the base: $9 - (- 1) = 10$ . The height to the third vertex is $6 - 1 = 5$ .
5Area $= \frac{1}{2} \times 10 \times 5 = 25$ . By the formula too: $\frac{1}{2} ∣ - 1 (1 - 6) + 9 (6 - 1) + 4 (1 - 1) ∣ = \frac{1}{2} ∣5 + 45 + 0∣ = 25$ ✓.

Answer Vertices $(4, 6)$ , $(- 1, 1)$ , $(9, 1)$ ; area 25 sq. units

A quick sketch shows the horizontal side at once.

Try it yourself

BasicTry it yourself first

Find the point of intersection of the lines

x + y = 5

and

x - y = 1

BasicTry it yourself first

Find the point of intersection of the lines

2 x + 3 y = 13

and

5 x - 2 y = 4

StandardTry it yourself first

Two straight roads on a map are the lines

3 x + 4 y = 2

and

5 x - 6 y = 16

. Find the coordinates of the crossing where they meet.

StandardTry it yourself first

For what value of

k

do the lines

k x + 2 y = 5

and

6 x + 4 y = 7

have no common point?

StandardTry it yourself first

The lines

x + 2 y = 10

3 x - y = 2

and

2 x + 5 y = k

pass through one point (are concurrent). Find

k

StandardTry it yourself first

Find the area of the triangle enclosed by the lines

x + y = 8

x - y = 0

and

y = 1

StandardTry it yourself first

Find the area of the triangle enclosed by the lines

x + y = 6

2 x - y = 0

and the X-axis.

AdvancedTry it yourself first

Find the area of the triangle enclosed by the lines

2 x + y = 4

x + 2 y = 5

and

x - y = 2

AdvancedTry it yourself first

Find the area of the triangle enclosed by the lines

3 x + 2 y = 12

x + 4 y = 8

and the Y-axis.

Parallel, perpendicular and the angle between lines

m_{1} = m_{2}

m_{1} m_{2} = - 1

tan θ

Parallel and perpendicular lines by comparing slopes, building their equations, and the angle between two lines.

The idea

The relation of two lines is decided by their slopes. Parallel lines lean equally the same way: equal slopes ( $m_{1} = m_{2}$ ). Perpendicular lines: turning one through $9 0^{\circ}$ gives the other: the step $(a, b) \to (- b, a)$ , so the slope $\frac{b}{a} \to - \frac{a}{b}$ , i.e. the reciprocal with a minus sign: $m_{1} m_{2} = - 1$ . The angle between any two lines comes from the difference of their slopes.

◭ Diagram

The line

2 x + y = 4

(black); through

P (3, 3)

the parallel

2 x + y = 9

(green) and the perpendicular

x - 2 y = - 3

(rose), which meets the black line at right angles at

(1, 2)

Relation	Slopes	Coefficients ( $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ )
Parallel	$m_{1} = m_{2}$	$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$
Perpendicular	$m_{1} m_{2} = - 1$	$a_{1} a_{2} + b_{1} b_{2} = 0$
Angle $θ$	$tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$

Here $m_{1}, m_{2}$ are the slopes of the two lines ( $m = - \frac{a}{b}$ ) and $θ$ is the acute angle between them. Rule for building lines: a parallel to $a x + b y = c$ is $a x + b y = c^{'}$ and a perpendicular is $b x - a y = c^{'}$ ; each time the new constant $c^{'}$ comes from putting in the given point.

◭ Diagram

The lines

y = 2 x - 1

and

y = \frac{x}{3} + 4

meet at

(3, 5)

; slopes

2

and

\frac{1}{3}

, so

tan θ = \frac{2 - \frac{1}{3}}{1 + \frac{2}{3}} = 1

, i.e.

θ = 4 5^{\circ}

Spot itRecognise it: “parallel”, “perpendicular”, “cut at right angles”, “angle between”, “value of

k

for which …”. Trap: for perpendicular the product is

- 1

, not

+ 1

; and if one line is vertical (

x = k

) the perpendicular is horizontal (

y = k^{'}

) — the slope formula will not work. In the angle formula the denominator has

1 + m_{1} m_{2}

+

sign).

Steps

Find the slopes of both lines ( $- \frac{a}{b}$ ).
Equal means parallel; product $- 1$ means perpendicular; otherwise the angle formula.
To build a parallel or perpendicular line, use the $c^{'}$ rule from $a x + b y = c$ and find $c^{'}$ from the point.
Check that both the point and the slope fit.

Solved examples

Example 13 · Basic

Find the equation of the line through

(3, - 2)

parallel to the line

4 x + 3 y = 7

Solution, step by step

1The parallel keeps the coefficients of $x$ and $y$ : $4 x + 3 y = c^{'}$ .
2The point $(3, - 2)$ lies on it: $c^{'} = 4 (3) + 3 (- 2) = 12 - 6 = 6$ .
3The line is $4 x + 3 y = 6$ . Check: slope $- \frac{4}{3}$ , same as before ✓.

Answer $4 x + 3 y = 6$

The parallel has

6

in place of

7

— only the constant changed.

Example 14 · Standard

Find the equation of the line through

(1, 4)

perpendicular to the line

2 x - 5 y = 3

Solution, step by step

1The slope of the given line is $- \frac{2}{- 5} = \frac{2}{5}$ . The perpendicular has the reciprocal with a minus sign: $- \frac{5}{2}$ .
2Point-slope: $y - 4 = - \frac{5}{2} (x - 1)$ . Multiply by 2: $2 y - 8 = - 5 x + 5$ .
3So $5 x + 2 y = 13$ .
4Check: $a_{1} a_{2} + b_{1} b_{2} = 2 \times 5 + (- 5) \times 2 = 0$ ✓ and the point: $5 (1) + 2 (4) = 13$ ✓.

Answer $5 x + 2 y = 13$

By the rule

b x - a y = c^{'}

directly:

- 5 x - 2 y = c^{'}

, the same line.

Example 15 · Advanced

Find the acute angle between the lines

3 x - y = 5

and

x - 2 y = - 4

Solution, step by step

1Slopes: for the first $m_{1} = - \frac{3}{- 1} = 3$ ; for the second $m_{2} = - \frac{1}{- 2} = \frac{1}{2}$ .
2 $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} = \frac{3 - \frac{1}{2}}{1 + 3 \times \frac{1}{2}}$ .
3The numerator $3 - \frac{1}{2} = \frac{5}{2}$ and the denominator $1 + \frac{3}{2} = \frac{5}{2}$ . So $tan θ = 1$ .
4 $tan θ = 1$ gives $θ = 4 5^{\circ}$ .
5Check: the inclinations are $tan^{- 1} 3 \approx 71.5 7^{\circ}$ and $tan^{- 1} \frac{1}{2} \approx 26.5 7^{\circ}$ ; their difference is $4 5^{\circ}$ ✓.

Answer $4 5^{\circ}$

If the denominator is

0

(

m_{1} m_{2} = - 1

) then

tan θ

is undefined and the angle is

9 0^{\circ}

Try it yourself

BasicTry it yourself first

Find the equation of the line through

(2, 7)

parallel to the line

3 x + y = 5

StandardTry it yourself first

Find the equation of the line through

(- 4, 1)

parallel to the line

5 x - 2 y = 11

BasicTry it yourself first

Find the equation of the line through

(0, 3)

perpendicular to the line

x + 2 y = 6

StandardTry it yourself first

Find the equation of the line through

(4, - 3)

perpendicular to the line

3 x - 4 y = 7

StandardTry it yourself first

For what value of

k

are the lines

2 x + 3 y = 6

and

6 x + k y = 5

perpendicular?

StandardTry it yourself first

For what value of

k

are the lines

3 x + 4 y = 12

and

9 x + k y = 5

parallel?

StandardTry it yourself first

Find

tan θ

for the acute angle

θ

between the lines

3 x - y = - 3

and

x - 2 y = 2

AdvancedTry it yourself first

Find the equation of the line through

(- 5, - 2)

perpendicular to the line

7 x - 3 y = 4

AdvancedTry it yourself first

Find

tan θ

for the acute angle

θ

between the lines

2 x - y = - 3

and

x + 3 y = - 3

AdvancedTry it yourself first

For what value of

k

are the lines

5 x - 4 y = 3

and

8 x + k y = 7

perpendicular?

Distance from a point to a line

\frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

The perpendicular distance of a point from a line, the distance between parallel lines, and which side of a line a point is on.

The idea

The distance of a point from a line is the shortest distance, the length of the perpendicular dropped on the line. To find it put the point into the equation of the line: $a x_{1} + b y_{1} + c$ . This number says how far the point is “off” the line, but mixed with the size of the coefficients. Dividing by $a^{2} + b^{2}$ gives the true length. The sign of the number says which side of the line the point is on.

◭ Diagram

The distance of

P (5, 5)

from the line

3 x + 4 y = 10

\frac{∣15 + 20 - 10∣}{5} = 5

; the step to the foot

F (2, 1)

(3, 4)

of length 5. Through

P

runs the parallel

3 x + 4 y = 35

(green).

∑ Formula

d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

Distance of the point

(x_{1}, y_{1})

from the line

a x + b y + c = 0

$d$: perpendicular distance
$a, b, c$: coefficients of the line $a x + b y + c = 0$ (constant on the left)
$x_{1}, y_{1}$: coordinates of the point

∑ Formula

d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}} (a x + b y + c_{1} = 0, a x + b y + c_{2} = 0), d_{O} = \frac{∣ c ∣}{a ^{2} + b ^{2}}

Distance between two parallel lines, and from the origin

$c_{1}, c_{2}$: the constants of the two lines (after making $a, b$ equal)
$d_{O}$: distance of the line from the origin

Spot itRecognise it: “distance of a point from a line”, “perpendicular distance”, “between parallel lines”, “radius of a circle touching the line”, “height”. Traps: (1) make the equation “

= 0

” first; (2) for parallel lines equalise

a, b

(turn

6 x + 8 y = 7

into

3 x + 4 y = 3.5

); (3) simplify the root in the denominator.

Steps

Write the line as $a x + b y + c = 0$ .
Put in the point and find the numerator $∣ a x_{1} + b y_{1} + c ∣$ .
Find the denominator $a^{2} + b^{2}$ and divide.
For parallel lines first equalise the coefficients, then use $∣ c_{1} - c_{2} ∣$ .

Solved examples

Example 16 · Basic

Find the distance of the point

(4, - 2)

from the line

3 x - 4 y = 5

Solution, step by step

1Make it “ $= 0$ ”: $3 x - 4 y - 5 = 0$ , i.e. $a = 3$ , $b = - 4$ , $c = - 5$ .
2Numerator: $∣3 (4) - 4 (- 2) - 5∣ = ∣12 + 8 - 5∣ = 15$ .
3Denominator: $3^{2} + (- 4)^{2} = 25 = 5$ .
4 $d = \frac{15}{5} = 3$ units.

Answer 3 units

The numerator is positive, so the point is on the side where

3 x - 4 y > 5

Example 17 · Standard

Find the distance between the parallel lines

5 x + 12 y = 10

and

10 x + 24 y = 7

Solution, step by step

1The formula needs equal coefficients of $x$ and $y$ . Divide the second line by 2: $5 x + 12 y = 3.5$ .
2Now $c_{1} = - 10$ and $c_{2} = - 3.5$ (in “ $= 0$ ” form). $∣ c_{1} - c_{2} ∣ = ∣ - 10 + 3.5∣ = 6.5$ .
3 $5^{2} + 1 2^{2} = 13$ .
4 $d = \frac{6.5}{13} = \frac{1}{2}$ unit.

Answer $\frac{1}{2}$ unit

Using

∣10 - 7∣/26

without equalising the coefficients is wrong.

Example 18 · Advanced

The distance of the point

(k, - 2)

from the line

3 x + 4 y = 7

is 3 units. Find all values of

k

Solution, step by step

1The line is $3 x + 4 y - 7 = 0$ . Distance formula: $\frac{∣3 k + 4 ( - 2 ) - 7∣}{9 + 16} = 3$ .
2The denominator is 5, so $∣3 k - 15∣ = 15$ .
3Two cases: $3 k - 15 = 15$ ⟹ $k = 10$ ; or $3 k - 15 = - 15$ ⟹ $k = 0$ .
4Check: $(10, - 2)$ : $∣30 - 8 - 7∣/5 = 3$ ✓; $(0, - 2)$ : $∣ - 8 - 7∣/5 = 3$ ✓.

Answer $k = 0$ or $k = 10$

The two points lie 3 units away on either side of the line.

Try it yourself

BasicTry it yourself first

Find the perpendicular distance of the point

(4, 5)

from the line

3 x + 4 y = 7

BasicTry it yourself first

Find the perpendicular distance of the point

(2, 3)

from the line

5 x + 12 y = 20

StandardTry it yourself first

Find the perpendicular distance of the point

(2, 3)

from the line

x + y = 1

StandardTry it yourself first

Find the distance between the parallel lines

3 x + 4 y = 10

and

6 x + 8 y = 50

StandardTry it yourself first

Find the distance between the parallel lines

5 x + 12 y = 6

and

15 x + 36 y = 96

StandardTry it yourself first

The distance of the point

(k, - 2)

from the line

3 x + 4 y = 7

is 6 units. Find all values of

k

StandardTry it yourself first

Do the points

A (3, 2)

and

B (- 1, 1)

lie together on one side of the line

2 x + y = 6

AdvancedTry it yourself first

Find the perpendicular distance of the point

(- 4, 3)

from the line

2 x - y = 7

AdvancedTry it yourself first

The distance of the point

(k, 1)

from the line

5 x + 12 y = 2

is 5 units. Find all values of

k

Foot and image

F = P - s (a, b)

P^{'} = 2 F - P

The foot of the perpendicular from a point to a line, the image in the line, and the mirror from a point and its image.

The idea

The perpendicular from $P$ to the line $a x + b y = c$ has direction $(a, b)$ (the normal direction of the line). So to reach the foot $F$ walk from $P$ in this direction as far as it takes to reach the line: $F = P - s (a, b)$ , where $s = \frac{a x _{1} + b y _{1} - c}{a ^{2} + b ^{2}}$ . The image $P^{'}$ is as far again: $P^{'} = P - 2 s (a, b) = 2 F - P$ , because $F$ is the midpoint of $P$ and $P^{'}$ .

◭ Diagram

From

P (5, 5)

3 x + 4 y = 10

: the foot is

F (2, 1)

and the image

P^{'} (- 1, - 3)

PF = F P^{'} = 5

, and

P P^{'}

is perpendicular to the line.

∑ Formula

s = \frac{a x _{1} + b y _{1} - c}{a ^{2} + b ^{2}}, F = (x_{1} - s a, y_{1} - s b), P^{'} = (x_{1} - 2 s a, y_{1} - 2 s b)

Point

P (x_{1}, y_{1})

, line

a x + b y = c

$s$: the multiplier that takes $P$ to the line
$F$: the foot of the perpendicular
$P^{'}$: the image
$a, b, c$: coefficients of the line $a x + b y = c$

The reverse question: given $P$ and its image $P^{'}$ , the mirror is the perpendicular bisector of $P P^{'}$ : it passes through the midpoint and is perpendicular to $P P^{'}$ . The same method gives the line of all points equidistant from two points (a locus).

Spot itRecognise it: “foot”, “foot of the perpendicular”, “image”, “reflection”, “equation of the mirror”, “a ray of light striking and bouncing”. Two checks: (1) the foot satisfies the equation of the line; (2)

PF = F P^{'}

and

P

F

P^{'}

are collinear.

Steps

Write the line as $a x + b y = c$ and find $s = \frac{a x _{1} + b y _{1} - c}{a ^{2} + b ^{2}}$ .
Foot: $F = (x_{1} - s a, y_{1} - s b)$ .
Image: $P^{'} = 2 F - P$ (or with $2 s$ ).
Check: the foot is on the line and the midpoint of $P P^{'}$ is the foot.

Solved examples

Example 19 · Basic

Find the foot of the perpendicular from the point

(- 2, 5)

to the line

x - y = 1

Solution, step by step

1The line has $a = 1$ , $b = - 1$ , $c = 1$ . Putting in the point gives $a x_{1} + b y_{1} - c = - 2 - 5 - 1 = - 8$ and $a^{2} + b^{2} = 2$ .
2 $s = \frac{- 8}{2} = - 4$ .
3Foot $= (x_{1} - s a, y_{1} - s b) = (- 2 + 4, 5 - 4) = (2, 1)$ .
4Check: $2 - 1 = 1$ ✓ and the direction of $PF$ is $(4, - 4)$ , proportional to $(1, - 1)$ (perpendicular) ✓.

Answer $(2, 1)$

s

came out negative because the point is on the side where

x - y < 1

Example 20 · Standard

Find the image of the point

(3, 8)

in the line

x + 3 y = 7

Solution, step by step

1 $a = 1$ , $b = 3$ , $c = 7$ . $a x_{1} + b y_{1} - c = 3 + 24 - 7 = 20$ and $a^{2} + b^{2} = 10$ , so $s = 2$ .
2Foot $F = (3 - 2 \times 1, 8 - 2 \times 3) = (1, 2)$ . Check: $1 + 6 = 7$ ✓.
3Image $= 2 F - P = (2 - 3, 4 - 8) = (- 1, - 4)$ .
4Check: the midpoint of $P$ and $P^{'}$ is $(1, 2)$ , on the line ✓; $P P^{'} = 16 + 144 = 160 = 410$ , which equals $2 \times \frac{20}{10} = 410$ ✓.

Answer $(- 1, - 4)$

Directly with

2 s

(3 - 4, 8 - 12)

Example 21 · Advanced

In a mirror the image of

(1, 2)

(3, 6)

. Find the equation of the mirror.

Solution, step by step

1The mirror is the perpendicular bisector of the segment joining the point and its image.
2The midpoint $(\frac{1 + 3}{2}, \frac{2 + 6}{2}) = (2, 4)$ is on the mirror.
3The segment has slope $\frac{6 - 2}{3 - 1} = 2$ , so the mirror has slope $- \frac{1}{2}$ .
4Point-slope: $y - 4 = - \frac{1}{2} (x - 2)$ , i.e. $2 y - 8 = - x + 2$ , i.e. $x + 2 y = 10$ .
5Check: the distance from $(1, 2)$ is $\frac{∣1 + 4 - 10∣}{5} = 5$ and from $(3, 6)$ is $\frac{∣3 + 12 - 10∣}{5} = 5$ ✓ (equal, on opposite sides).

Answer $x + 2 y = 10$

This line is also the locus of points equidistant from

(1, 2)

and

(3, 6)

Try it yourself

BasicTry it yourself first

Find the foot of the perpendicular from the point

(4, 6)

to the line

x + y = 4

BasicTry it yourself first

Find the foot of the perpendicular from the point

(1, 0)

to the line

x - y = 3

StandardTry it yourself first

Find the image of the point

(5, 5)

in the line

4 x + 3 y = 10

StandardTry it yourself first

Find the image of the point

(0, 6)

in the line

x + y = 2

StandardTry it yourself first

In a mirror the image of the point

(2, 1)

(6, 5)

. Find the equation of the mirror line.

StandardTry it yourself first

Find the foot of the perpendicular from the point

(9, 8)

to the line

4 x + 3 y = 10

StandardTry it yourself first

Find the distance between the point

(2, 6)

and its image in the line

3 x + 4 y = 5

AdvancedTry it yourself first

In a mirror the image of the point

(- 1, 3)

(5, - 1)

. Find the equation of the mirror line.

AdvancedTry it yourself first

Find the image of the point

(2, - 5)

in the line

2 x - 3 y = 6

Triangle lines, regions, modulus

Median · inequality ·

∣ x ∣

Equations of special lines of a triangle, regions of inequalities and shapes with $∣ x ∣$ , $∣ y ∣$ .

Putting the methods above together, three common exam types: equations of the median, altitude and perpendicular bisector of a triangle; the shaded region of an inequality; and areas of shapes with modulus.

Median, altitude, perpendicular bisector

Each line = two pieces of information

Equations of the special lines of a triangle, the circumcentre and orthocentre, and a line through the meeting of two lines.

The idea

To build the equation of a line we need two pieces of information (two points, or a point and a slope). The three special lines of a triangle fit these. Median: the vertex and the midpoint of the opposite side, i.e. two points. Altitude: the vertex and the direction perpendicular to the opposite side, i.e. a point and a slope. Perpendicular bisector: the midpoint of a side and the direction perpendicular to it, i.e. a point and a slope. The tools from the previous chapter and methods simply combine here.

◭ Diagram

Triangle

A (1, 1)

B (9, 3)

C (3, 7)

: the median

A M

(

M

the midpoint of

BC

, green), the altitude on

BC

(rose, with the right-angle mark) and the perpendicular bisector of

A B

(violet).

Line	Through	Direction	Method
Median (vertex $A$ )	$A$	towards the midpoint $M$ opposite	two-point form
Altitude ( $A$ onto $BC$ )	$A$	perpendicular to $BC$	$b x - a y = c^{'}$ rule
Perpendicular bisector (of $A B$ )	midpoint of $A B$	perpendicular to $A B$	$b x - a y = c^{'}$ rule

The three altitudes meet at the orthocentre $H$ and the three perpendicular bisectors at the circumcentre $O$ . We only need the meeting of two lines. One more tool: any line through the intersection point of $L_{1} = 0$ and $L_{2} = 0$ is $L_{1} + λ L_{2} = 0$ ; $λ$ is fixed by one more condition (a point, a slope). It saves finding the meeting point.

Spot itRecognise it: “equation of the median”, “equation of the altitude”, “perpendicular bisector”, “coordinates of the circumcentre/orthocentre”, “line through the point of intersection”. Care: for an altitude use the direction of the opposite side

BC

; and the midpoint for a median is on the opposite side.

Steps

Decide which line it is and which two pieces of information it needs.
Find the midpoint or the direction needed.
Build the equation by the two-point form or the perpendicular rule and check it.
If a centre is asked, solve two such lines.

Solved examples

Example 22 · Basic

In the triangle

A (0, 4)

B (6, 0)

C (2, - 2)

find the equation of the median from

A

Solution, step by step

1The median joins $A$ to the midpoint of $BC$ : $M = (\frac{6 + 2}{2}, \frac{0 - 2}{2}) = (4, - 1)$ .
2The slope through $A (0, 4)$ and $M (4, - 1)$ is $\frac{- 1 - 4}{4 - 0} = - \frac{5}{4}$ .
3With $A$ on the Y-axis: $y = - \frac{5}{4} x + 4$ , i.e. $4 y = - 5 x + 16$ , i.e. $5 x + 4 y = 16$ .
4Check: at $M$ , $5 (4) + 4 (- 1) = 20 - 4 = 16$ ✓.

Answer $5 x + 4 y = 16$

A

has abscissa 0, so

c

came directly as 4.

Example 23 · Standard

In the same triangle find the altitude from

C

A B

and the perpendicular bisector of

A B

Solution, step by step

1The direction of $A B$ is $B - A = (6, - 4)$ . Both lines are perpendicular to $A B$ , so both have $(a, b) = (6, - 4)$ , i.e. $6 x - 4 y = c$ (or simply $3 x - 2 y = c^{'}$ ).
2The altitude passes through $C (2, - 2)$ : $c^{'} = 3 (2) - 2 (- 2) = 6 + 4 = 10$ , so $3 x - 2 y = 10$ .
3The perpendicular bisector passes through the midpoint of $A B$ , $(3, 2)$ : $c^{'} = 3 (3) - 2 (2) = 9 - 4 = 5$ , so $3 x - 2 y = 5$ .
4The two are parallel (same $3 x - 2 y$ ) — as they must be, both being perpendicular to $A B$ . Check: the slope of $A B$ is $- \frac{2}{3}$ and theirs is $\frac{3}{2}$ , product $- 1$ ✓.

Answer Altitude $3 x - 2 y = 10$ ; perpendicular bisector $3 x - 2 y = 5$

The altitude goes through the vertex, the bisector through the midpoint — that is the only difference.

Example 24 · Advanced

Find the circumcentre of the triangle

A (4, 2)

B (- 3, 1)

C (6, - 2)

using perpendicular bisectors.

Solution, step by step

1 $A B$ : midpoint $(0.5, 1.5)$ and direction $(- 7, - 1)$ . The perpendicular bisector is $7 x + y = c$ with $c = 7 (0.5) + 1.5 = 5$ , i.e. $7 x + y = 5$ .
2 $BC$ : midpoint $(1.5, - 0.5)$ and direction $(9, - 3)$ , simplified $(3, - 1)$ . The perpendicular bisector is $3 x - y = c$ with $c = 3 (1.5) + 0.5 = 5$ , i.e. $3 x - y = 5$ .
3Adding gives $10 x = 10$ , so $x = 1$ ; then $y = 3 (1) - 5 = - 2$ . The circumcentre is $(1, - 2)$ .
4Check: $O A^{2} = 9 + 16 = 25$ , $O B^{2} = 16 + 9 = 25$ , $O C^{2} = 25 + 0 = 25$ — all equal, radius 5 ✓.

Answer $(1, - 2)$ , circumradius 5

The third bisector (of

C A

) must also pass through this point — a handy check.

Try it yourself

BasicTry it yourself first

The vertices of triangle

A BC

are

A (0, 0)

B (6, 4)

C (2, 10)

. Find the equation of the median from

A

BasicTry it yourself first

The vertices of triangle

A BC

are

A (1, 5)

B (- 2, 2)

C (6, 4)

. Find the equation of the altitude from

A

BC

BasicTry it yourself first

Find the perpendicular bisector of the segment

A B

, where

A (1, 2)

and

B (5, 6)

StandardTry it yourself first

The vertices of triangle

A BC

are

A (3, - 4)

B (7, 6)

C (1, 2)

. Find the equation of the median from

A

StandardTry it yourself first

The vertices of triangle

A BC

are

A (- 3, 2)

B (1, - 1)

C (5, 5)

. Find the equation of the altitude from

A

BC

StandardTry it yourself first

Find the perpendicular bisector of the segment

A B

, where

A (- 2, 7)

and

B (4, 1)

StandardTry it yourself first

The vertices of a triangle are

A (- 2, 4)

B (5, 5)

C (6, - 2)

. Using perpendicular bisectors, find the coordinates of its circumcentre.

StandardTry it yourself first

Find the equation of the line through the point of intersection of

x + y = 6

and

2 x - y = 3

and through the point

(4, 0)

AdvancedTry it yourself first

The vertices of a triangle are

A (- 3, - 3)

B (- 1, 1)

C (5, 1)

. Find its orthocentre.

AdvancedTry it yourself first

Find the equation of the line through the point of intersection of

3 x + 2 y = 7

and

x - y = 1

and through the point

(0, 5)

Regions of linear inequalities

Origin test · shading

The half-plane of $a x + b y \leq c$ , deciding the shading, the common region of several inequalities and its area.

The idea

The line $a x + b y = c$ splits the plane into two half-planes. On one $a x + b y < c$ and on the other $a x + b y > c$ . To know which side the inequality is true on, put in one test point: the easiest is the origin $(0, 0)$ (provided the line does not pass through it). If it satisfies the inequality choose its side, otherwise the other side. With “ $\leq$ / $\geq$ ” the line belongs to the region (solid line), with “ $<$ / $>$ ” it does not (dashed line).

◭ Diagram

For

3 x + 2 y \leq 12

: at the origin

0 \leq 12

is true, so the shading (green) is on the side with the origin; on the other side

3 x + 2 y > 12

If the line passes through the origin ( $c = 0$ ) the origin test fails; take another easy point, such as $(1, 0)$ or $(0, 1)$ . With several inequalities find each region and take the common part; it is a polygon whose vertices are meetings of the lines, with the area from the previous chapter’s formula. For “how many integer points” count row by row and add.

◭ Diagram

The common region of

x \geq 0

y \geq 0

x \leq 4

x + y \leq 6

: vertices

(0, 0)

(4, 0)

(4, 2)

(0, 6)

; area

\frac{1}{2} (6 + 2) \times 4 = 16

Spot itRecognise it: “region of the inequality”, “shaded part”, “points satisfying”, “area of the common region”, “how many integer points”. Traps: (1) if the sign flips (

\geq

instead of

\leq

) the side changes; (2)

x \geq 0

y \geq 0

means the first quadrant; (3) in a strict inequality the points on the line do not count.

Steps

Draw the boundary line $a x + b y = c$ of each inequality (using intercepts).
Test the origin (or another point) and pick the correct side.
Find the common part and its vertices.
Then find the area or count the integer points as asked.

Solved examples

Example 25 · Basic

On which side is the shaded region of

3 x + 5 y \geq 30

? Is the point

(6, 4)

in it?

Solution, step by step

1The boundary is $3 x + 5 y = 30$ , which cuts the axes at $(10, 0)$ and $(0, 6)$ . The origin is not on it.
2Test: at $(0, 0)$ the statement $0 \geq 30$ is false, so the shading is away from the origin. The line is solid ( $\geq$ ).
3At $(6, 4)$ , $3 (6) + 5 (4) = 18 + 20 = 38 \geq 30$ is true, so it is in the region.

Answer Away from the origin; yes, $(6, 4)$ is in the region

Any test point gives the same verdict, since a whole half-plane is true or false together.

Example 26 · Standard

Determine the region of

x \geq 2 y

(the line passes through the origin). Which of

(1, 3)

and

(5, 1)

lies in it?

Solution, step by step

1The boundary $x = 2 y$ passes through the origin, so the origin test is useless ( $0 \geq 0$ is true only because it is on the line).
2Take another point, say $(1, 0)$ : $1 \geq 0$ is true, so the shading is on the side of $(1, 0)$ , below and to the right of the line.
3 $(1, 3)$ : $1 \geq 6$ is false. $(5, 1)$ : $5 \geq 2$ is true.

Answer $(5, 1)$ is in the region; $(1, 3)$ is not

For a line through the origin take

(1, 0)

(0, 1)

, whichever is not on the line.

Example 27 · Advanced

The common region of

x \geq 0

y \geq 0

x + y \leq 7

and

x + 2 y \geq 4

is a polygon. Find its vertices and area.

Solution, step by step

1The first three give the big triangle $(0, 0)$ , $(7, 0)$ , $(0, 7)$ . The fourth, $x + 2 y \geq 4$ , cuts off the origin corner, since $0 \geq 4$ is false; the shading is above and to the right of the line $x + 2 y = 4$ .
2The line $x + 2 y = 4$ cuts the axes at $(4, 0)$ and $(0, 2)$ . So the vertices are $(4, 0)$ , $(7, 0)$ , $(0, 7)$ , $(0, 2)$ .
3Area = big triangle $-$ cut-off triangle $= \frac{1}{2} \times 7 \times 7 - \frac{1}{2} \times 4 \times 2 = 24.5 - 4 = 20.5$ .
4Shoelace check ( $(4, 0), (7, 0), (0, 7), (0, 2)$ ): $4 \cdot 0 - 7 \cdot 0 = 0$ , $7 \cdot 7 - 0 \cdot 0 = 49$ , $0 \cdot 2 - 0 \cdot 7 = 0$ , $0 \cdot 0 - 4 \cdot 2 = - 8$ ; sum $41$ , half $20.5$ ✓.

Answer Vertices $(4, 0)$ , $(7, 0)$ , $(0, 7)$ , $(0, 2)$ ; area $20.5$ sq. units

“Big minus small” is often faster than the shoelace.

Try it yourself

BasicTry it yourself first

Which of these points satisfies the inequality

2 x + 3 y \leq 12

BasicTry it yourself first

How is the region of the inequality

4 x + 3 y \leq 24

shaded in a diagram?

StandardTry it yourself first

How is the region of the inequality

5 x - 2 y > 10

shaded in a diagram?

StandardTry it yourself first

Find the area of the region given by

x \geq 0

y \geq 0

and

2 x + 3 y \leq 12

StandardTry it yourself first

Which of these points satisfies the inequality

3 x - 2 y \geq 6

StandardTry it yourself first

Find the area of the region given by

x \geq 0

y \geq 0

x \leq 3

and

x + y \leq 8

StandardTry it yourself first

How many points

(x, y)

with integer coordinates satisfy

x \geq 0

y \geq 0

and

x + y \leq 6

AdvancedTry it yourself first

How many points

(x, y)

with integer coordinates satisfy

x \geq 0

y \geq 0

and

2 x + 3 y \leq 12

AdvancedTry it yourself first

Find the area of the region given by

x \geq 0

y \geq 0

x \leq 5

and

x + y \leq 9

Shapes with modulus ( $∣ x ∣$ , $∣ y ∣$ )

∣ x ∣ + ∣ y ∣ = k \Rightarrow 2 k^{2}

$∣ x ∣ + ∣ y ∣ = k$ , $\frac{∣ x ∣}{a} + \frac{∣ y ∣}{b} = 1$ , shifted figures, rectangles and counting integer points.

The idea

$∣ x ∣$ means $x$ without its sign: $x$ when $x \geq 0$ , and $- x$ when $x < 0$ . So when an equation has $∣ x ∣$ or $∣ y ∣$ we get a different line in each quadrant, and four pieces join to make the figure. In $∣ x ∣ + ∣ y ∣ = k$ each piece is $\pm x \pm y = k$ , cutting the axes at $\pm k$ : the figure is a square whose diagonals lie on the axes. It shows that these areas are 4 times the triangle in the first quadrant.

◭ Diagram

\frac{∣ x ∣}{3} + \frac{∣ y ∣}{5} = 1

: corners

(\pm 3, 0)

and

(0, \pm 5)

; a rhombus with diagonals 6 and 10, area

\frac{1}{2} \times 6 \times 10 = 30 = 2 ab

Equation / inequality	Shape	Area
$∣ x ∣ + ∣ y ∣ = k$	square	$2 k^{2}$
$\frac{∣ x ∣}{a} + \frac{∣ y ∣}{b} = 1$	rhombus	$2 ab$
$∣ x - h ∣ + ∣ y - j ∣ = r$	the same square, centre $(h, j)$	$2 r^{2}$
$∣ x ∣ \leq a, ∣ y ∣ \leq b$	rectangle	$4 ab$
$x + ∣ y ∣ = k$ and the Y-axis	triangle	$k^{2}$

Here $k$ is the “radius” of the square (the cut on an axis), $a, b$ the cuts on the axes, $(h, j)$ the centre of the figure and $r$ the distance from the centre to a corner (along an axis).

◭ Diagram

x + ∣ y ∣ = 4

becomes

x + y = 4

above and

x - y = 4

below; with the Y-axis the triangle

(0, 4)

(0, - 4)

(4, 0)

has area

\frac{1}{2} \times 8 \times 4 = 16 = k^{2}

Spot itRecognise it: a

∣ ∣

in the equation. The quickest way: find the cuts on the axes (put

y = 0

for

x

x = 0

for

y

), recognise the figure and use its diagonals. Traps: (1) the perimeter of

∣ x ∣ + ∣ y ∣ = k

4 k 2

(side

k 2

), not

4 k

; (2) a shifted figure keeps its area; (3) count the boundary points in lattice counts (

\leq

Steps

Find the cuts on the axes (put $y = 0$ and $x = 0$ ).
Recognise the figure: rhombus, square, rectangle or triangle.
Find the area or perimeter from the diagonals/sides.
For a shifted figure first put $X = x - h$ , $Y = y - j$ to bring the centre to the origin.

Solved examples

Example 28 · Basic

Find the area and the perimeter of the figure enclosed by

∣ x ∣ + ∣ y ∣ = 6

Solution, step by step

1At $y = 0$ , $∣ x ∣ = 6$ , so $x = \pm 6$ ; at $x = 0$ , $y = \pm 6$ . The corners are $(\pm 6, 0)$ and $(0, \pm 6)$ .
2Both diagonals are 12 and perpendicular, so it is a square.
3Area $= \frac{1}{2} \times 12 \times 12 = 72$ .
4Side $= 6^{2} + 6^{2} = 62$ , so the perimeter is $4 \times 62 = 242$ .

Answer Area 72, perimeter $242$

The side is

k 2

, not

k

Example 29 · Standard

Find the area, perimeter and side of the figure enclosed by

\frac{∣ x ∣}{4} + \frac{∣ y ∣}{3} = 1

Solution, step by step

1Corners $(\pm 4, 0)$ and $(0, \pm 3)$ . Diagonals $8$ and $6$ .
2It is a rhombus: area $= \frac{1}{2} \times 8 \times 6 = 24$ (formula $2 ab = 2 \times 4 \times 3$ ).
3A side runs from $(4, 0)$ to $(0, 3)$ : $16 + 9 = 5$ . The perimeter is $4 \times 5 = 20$ .
4Check: the origin is $\frac{12}{5} = 2.4$ from the side $3 x + 4 y = 12$ ; the height of the rhombus is twice that, $4.8$ , and side $\times$ height $= 5 \times 4.8 = 24$ ✓.

Answer Area 24, side 5, perimeter 20

The

(3, 4, 5)

triple makes the side a whole number.

Example 30 · Advanced

How many points with integer coordinates satisfy

∣ x ∣ + ∣ y ∣ \leq 5

Solution, step by step

1The integer values of $x$ run from $- 5$ to $5$ . For each $x$ , $∣ y ∣ \leq 5 - ∣ x ∣$ , i.e. $2 (5 - ∣ x ∣) + 1$ integer values of $y$ .
2 $x = 0$ : $11$ ; $x = \pm 1$ : $9$ each; $x = \pm 2$ : $7$ each; $x = \pm 3$ : $5$ each; $x = \pm 4$ : $3$ each; $x = \pm 5$ : $1$ each.
3Total $= 11 + 2 (9 + 7 + 5 + 3 + 1) = 11 + 50 = 61$ .
4Formula check: $2 n^{2} + 2 n + 1 = 2 (25) + 10 + 1 = 61$ ✓.

Answer 61

The

20

boundary points (

∣ x ∣ + ∣ y ∣ = 5

) are included;

41

are strictly inside.

Try it yourself

BasicTry it yourself first

Find the area of the figure enclosed by

∣ x ∣ + ∣ y ∣ = 7

BasicTry it yourself first

Find the area enclosed by

\frac{∣ x ∣}{6} + \frac{∣ y ∣}{8} = 1

StandardTry it yourself first

Find the area enclosed by

∣ x - 3∣ + ∣ y + 2∣ = 5

StandardTry it yourself first

Find the perimeter of the figure enclosed by

\frac{∣ x ∣}{5} + \frac{∣ y ∣}{12} = 1

StandardTry it yourself first

Find the area of the region given by

∣ x ∣ \leq 7

and

∣ y ∣ \leq 4

StandardTry it yourself first

Find the area of the region enclosed by

x + ∣ y ∣ = 9

and the Y-axis.

StandardTry it yourself first

How many points

(x, y)

with integer coordinates satisfy

∣ x ∣ + ∣ y ∣ \leq 3

AdvancedTry it yourself first

Find the area of the figure enclosed by

∣ x ∣ + ∣ y ∣ = 11

AdvancedTry it yourself first

How many points

(x, y)

with integer coordinates satisfy

∣ x ∣ + ∣ y ∣ \leq 6

Locus and the circle

A figure from a rule · the circle

The path of a point obeying a rule, and the equation of a circle.

Here we turn things round: no figure is given, only a rule, and we find the equation of the figure (a line or a circle).

Locus and the equation of a circle

Rule

\to

equation ·

(x - h)^{2} + (y - k)^{2} = r^{2}

The locus of equal distance, fixed distance from a line, the equation of a circle, centre and radius, the circle on a diameter and tangency.

The idea

A locus is the set of all points obeying a given rule. There is one way to find its equation: name the moving point $P (x, y)$ , write the rule in terms of $x$ and $y$ (mostly with the distance formula) and simplify. The equation that is left is the locus. Three rules come again and again: equal distance from two points (result: the perpendicular bisector, a line), fixed distance from a point (result: a circle), and fixed distance from a line (result: two parallel lines).

◭ Diagram

All points equally far from

A (1, 2)

and

B (5, - 2)

lie on the line

x - y = 3

(the perpendicular bisector of

A B

). For

P (6, 3)

P A = PB = 26

; the moving dot runs along the whole line.

◭ Diagram

Points 3 units from the line

3 x + 4 y = 7

lie on the two parallels

3 x + 4 y = 22

and

3 x + 4 y = - 8

(change

\pm 3 \times 5 = \pm 15

A circle is all points at distance $r$ from the centre $(h, k)$ . Squaring $PC = r$ in the distance formula gives $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Expanding gives the general form $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ , in which the centre is $(- g, - f)$ and the radius is $g^{2} + f^{2} - c$ . The key to recognising it: the coefficients of $x^{2}$ and $y^{2}$ are equal and there is no $x y$ term.

◭ Diagram

The circle

(x - 3)^{2} + (y - 2)^{2} = 25

: centre

C (3, 2)

, radius 5; at

(7, 5)

16 + 9 = 5

✓. The moving dot stays 5 from the centre all the time.

∑ Formula

(x - h)^{2} + (y - k)^{2} = r^{2}, x^{2} + y^{2} + 2 gx + 2 f y + c = 0 \Rightarrow centre (- g, - f), r = g^{2} + f^{2} - c

Both forms of the circle

$(h, k)$: the centre
$r$: the radius
$g, f, c$: constants of the general form ( $2 g$ and $2 f$ are the coefficients of $x$ and $y$ )

Situation	Result
diameter ends $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$	$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$
touches a line	radius = distance from centre to the line
touches both axes in the first quadrant	centre $(r, r)$
point $P$ inside / on / outside	$P C^{2} < r^{2}$ / $= r^{2}$ / $> r^{2}$

Spot itRecognise it: “locus”, “moving point”, “equidistant”, “constant distance”, “equation of a circle”, “centre and radius”, “ends of a diameter”, “touches the line”. Traps: (1) if the coefficient of

x^{2}

is not 1, divide first; (2) the centre’s sign looks reversed in

(x - h)

(x + 2)^{2}

means

h = - 2

; (3) the equation gives

r^{2}

; take the root for

r

Steps

Call the moving point $P (x, y)$ and say the rule in words.
Write the rule with the distance formula (or another formula) in $x$ , $y$ ; square to remove roots.
Expand and simplify: if $x^{2}$ , $y^{2}$ cancel it is a line; if they stay (with equal coefficients) it is a circle.
Check with one point.

Solved examples

Example 31 · Basic

A point

P

moves so that it is equally far from

A (1, 2)

and

B (5, - 2)

. Find the locus of

P

Solution, step by step

1Take $P (x, y)$ . The rule $P A = PB$ gives $P A^{2} = P B^{2}$ .
2 $(x - 1)^{2} + (y - 2)^{2} = (x - 5)^{2} + (y + 2)^{2}$ .
3Expand: $x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = x^{2} - 10 x + 25 + y^{2} + 4 y + 4$ .
4 $x^{2}$ and $y^{2}$ cancel: $- 2 x - 4 y + 5 = - 10 x + 4 y + 29$ , i.e. $8 x - 8 y = 24$ , i.e. $x - y = 3$ .
5Check: the midpoint of $A B$ , $(3, 0)$ , is on it ( $3 - 0 = 3$ ✓) and $A B$ has slope $- 1$ while this line has slope $1$ — product $- 1$ ✓.

Answer $x - y = 3$

With equal distances

x^{2}

and

y^{2}

always cancel, so a line results.

Example 32 · Standard

(i) Find the centre and radius of the circle

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

. (ii) Write the equation of the circle with centre

(- 2, 3)

and radius 4.

Solution, step by step

1(i) $2 g = - 6$ , $2 f = 4$ , $c = - 12$ , so $g = - 3$ , $f = 2$ .
2Centre $(- g, - f) = (3, - 2)$ . Radius $= 9 + 4 - (- 12) = 25 = 5$ .
3(ii) $(x - (- 2))^{2} + (y - 3)^{2} = 4^{2}$ , i.e. $(x + 2)^{2} + (y - 3)^{2} = 16$ .
4Expand: $x^{2} + 4 x + 4 + y^{2} - 6 y + 9 = 16$ , i.e. $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ .
5Check (i): the point $(8, - 2)$ is 5 from $(3, - 2)$ and $64 + 4 - 48 - 8 - 12 = 0$ ✓ (the point is on the circle).

Answer (i) centre $(3, - 2)$ , $r = 5$ ; (ii) $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$

The centre’s signs are the reverse of

g

and

f

Example 33 · Advanced

The points

A (- 1, 2)

and

B (5, - 6)

are the ends of a diameter of a circle. Find its equation, centre and radius, and say whether the origin is inside or outside.

Solution, step by step

1The diameter formula: $(x + 1) (x - 5) + (y - 2) (y + 6) = 0$ .
2Expand: $x^{2} - 4 x - 5 + y^{2} + 4 y - 12 = 0$ , i.e. $x^{2} + y^{2} - 4 x + 4 y - 17 = 0$ .
3Centre: $(- g, - f) = (2, - 2)$ (the midpoint of $A B$ ). Radius $= 4 + 4 + 17 = 5$ . Check: $A B = 36 + 64 = 10 = 2 r$ ✓.
4The square of the origin’s distance from the centre is $4 + 4 = 8$ , less than $r^{2} = 25$ , so the origin is inside the circle.

Answer $x^{2} + y^{2} - 4 x + 4 y - 17 = 0$ ; centre $(2, - 2)$ , $r = 5$ ; the origin is inside

Putting

(0, 0)

in the general form gives

c = - 17 < 0

, a sign of “inside”.

Try it yourself

BasicTry it yourself first

A point

P (x, y)

moves so that it is always equally far from

A (0, 3)

and

B (6, - 1)

. Find the locus of

P

BasicTry it yourself first

Find the equation of the circle with centre

(2, - 3)

and radius 5.

BasicTry it yourself first

Find the centre and radius of the circle

x^{2} + y^{2} - 8 x + 2 y - 19 = 0

StandardTry it yourself first

A point

P (x, y)

moves so that it is always equally far from

A (- 3, 2)

and

B (1, 8)

. Find the locus of

P

StandardTry it yourself first

Find the equation of the circle with the points

A (2, 3)

and

B (8, - 5)

as the ends of a diameter.

StandardTry it yourself first

Where is the point

(5, 5)

with respect to the circle

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

StandardTry it yourself first

A circle with centre

(2, 1)

touches the line

3 x + 4 y = - 5

. Find its radius.

StandardTry it yourself first

Find the locus of all points at a distance of 2 units from the line

5 x + 12 y = 8

AdvancedTry it yourself first

Find the equation of the circle with centre

(- 4, 5)

and radius 7.

AdvancedTry it yourself first

Where is the point

(3, - 1)

with respect to the circle

x^{2} + y^{2} + 4 x - 6 y - 23 = 0

Summary, mixed practice, revision

Every method in one place

A one-page table of all methods, exam-style mixed questions and revision.

Summary of all methods

When · what · how

Every method: how to recognise the question and which formula to use.

Method	Recognise	Key formula / rule
Slope	slope, inclination, angle with the X-axis	$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = tan θ = - \frac{a}{b}$
Forms of the equation	build / read the equation of a line	$y = m x + c$ , $y - y_{1} = m (x - x_{1})$ , $\frac{x}{a} + \frac{y}{b} = 1$
Triangle with the axes	area enclosed by a line and the axes	$\frac{c ^{2}}{2∣ ab ∣}$ ; trapezium: $\frac{c _{2}^{2} - c _{1}^{2}}{2∣ ab ∣}$
Meeting of two lines	intersection, concurrent, triangle of three lines	eliminate; $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ parallel
Parallel, perpendicular, angle	build a parallel/perpendicular, angle	$m_{1} = m_{2}$ ; $m_{1} m_{2} = - 1$ ; $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$
Distance point to line	perpendicular distance, between parallels	$\frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$ ; $\frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}}$
Foot and image	foot of perpendicular, mirror image	$F = P - s (a, b)$ , $P^{'} = 2 F - P$ , $s = \frac{a x _{1} + b y _{1} - c}{a ^{2} + b ^{2}}$
Median, altitude, perp. bisector	equations of triangle lines, circumcentre, orthocentre	two-point; $b x - a y = c^{'}$ ; $L_{1} + λ L_{2} = 0$
Region of an inequality	shaded region, common region, integer points	origin test; area from vertices
Modulus shapes	an equation with $∣ x ∣$ , $∣ y ∣$	$∣ x ∣ + ∣ y ∣ = k : 2 k^{2}$ ; $\frac{∣ x ∣}{a} + \frac{∣ y ∣}{b} = 1 : 2 ab$ ; $x + ∣ y ∣ = k : k^{2}$
Locus, circle	equal/fixed distance, equation of a circle	$(x - h)^{2} + (y - k)^{2} = r^{2}$ ; $- g, - f$ ; $g^{2} + f^{2} - c$

Letters in the table: $m$ = slope, $θ$ = angle from the positive X-axis; $a, b, c$ = coefficients of the line $a x + b y = c$ ; $c_{1}, c_{2}$ = constants of parallel lines; $(x_{1}, y_{1})$ = the given point; in $x / a$ and $y / b$ , $a, b$ are the cuts on the axes; $F$ = the foot, $P^{'}$ = the image, $s$ = the multiplier; $λ$ = the family parameter; $k$ = the size of a modulus figure (cut on an axis); $(h, k)$ and $r$ = the centre and radius of a circle; $g, f$ = constants of the general form of a circle.

RememberIn every question ask: (1) is a line given or to be built (pick the form of the equation)? (2) does it involve the axes (cuts, area)? (3) is there a relation between two lines or a point and a line (compare slopes, the distance formula)? And at the end check each answer in the given points or lines.

Mixed practice

Exam-style questions · all methods

Mixed questions on all methods; first decide which method fits.

The questions here are not in the order of the methods. Before each one pause and decide which method applies. Open a solution only after you have tried.

BasicTry it yourself first

Find the slope of the line through

(2, - 5)

and

(6, 7)

StandardTry it yourself first

A line has slope

3

. What is its angle of inclination

θ

(measured anticlockwise from the positive X-axis)?

StandardTry it yourself first

A line makes an angle of

13 5^{\circ}

with the positive X-axis. Find its slope.

AdvancedTry it yourself first

The line through

(- 2, 4)

and

(k, - 5)

has slope

- 3

. Find

k

StandardTry it yourself first

Find the slope of the line

9 x - 6 y = 18

StandardTry it yourself first

Find the equation of the line through

(- 2, - 5)

with slope

\frac{3}{4}

AdvancedTry it yourself first

Find the equation of the line through

(2, - 3)

and

(- 4, 9)

StandardTry it yourself first

A line cuts the X-axis at

8

and the Y-axis at

- 5

. Find its equation.

StandardTry it yourself first

Find the Y-intercept of the line

9 x + 4 y = 36

StandardTry it yourself first

The point

(5, k)

lies on the line

4 x + 5 y = 40

. Find

k

StandardTry it yourself first

Find the area (sq. units) of the triangle enclosed by the line

6 x + 5 y = 30

and the two coordinate axes.

AdvancedTry it yourself first

Find the area of the trapezium enclosed by the lines

2 x + 3 y = 6

2 x + 3 y = 18

and the two axes.

StandardTry it yourself first

Find the perimeter of the triangle enclosed by the line

\frac{x}{9} + \frac{y}{12} = 1

and the two axes.

AdvancedTry it yourself first

The line

3 x + 5 y = c

and the two axes enclose a triangle of area 30 sq. units (

c > 0

). Find

c

StandardTry it yourself first

Find the point of intersection of the lines

4 x + 3 y = 14

and

x - 2 y = 9

AdvancedTry it yourself first

For what value of

k

do the lines

k x + 3 y = 7

and

8 x + 6 y = 9

have no common point?

StandardTry it yourself first

The lines

3 x + y = 10

x - y = 2

and

4 x + 3 y = k

pass through one point (are concurrent). Find

k

StandardTry it yourself first

Find the area of the triangle enclosed by the lines

x + y = 10

2 x - y = 2

and

y = 2

StandardTry it yourself first

Find the area of the triangle enclosed by the lines

x + 2 y = 12

x - y = 3

and the X-axis.

StandardTry it yourself first

Find the equation of the line through

(3, - 2)

parallel to the line

2 x + 5 y = 9

StandardTry it yourself first

Find the equation of the line through

(- 1, 6)

perpendicular to the line

4 x - 3 y = 5

StandardTry it yourself first

For what value of

k

are the lines

3 x - 4 y = 12

and

8 x + k y = 9

perpendicular?

AdvancedTry it yourself first

For what value of

k

are the lines

2 x - 5 y = 10

and

6 x + k y = 9

parallel?

AdvancedTry it yourself first

Find

tan θ

for the acute angle

θ

between the lines

3 x - y = - 3

and

x + 2 y = - 2

StandardTry it yourself first

Find the perpendicular distance of the point

(- 2, 3)

from the line

x - y = 3

StandardTry it yourself first

Find the distance between the parallel lines

4 x + 3 y = 5

and

12 x + 9 y = 90

AdvancedTry it yourself first

The distance of the point

(k, - 2)

from the line

5 x + 12 y = - 4

is 5 units. Find all values of

k

StandardTry it yourself first

Do the points

A (1, - 4)

and

B (2, 5)

lie together on one side of the line

3 x - y = 2

StandardTry it yourself first

Find the foot of the perpendicular from the point

(5, 5)

to the line

2 x + y = 5

AdvancedTry it yourself first

Find the image of the point

(2, 6)

in the line

x + 2 y = 4

AdvancedTry it yourself first

In a mirror the image of the point

(2, - 3)

(- 4, 5)

. Find the equation of the mirror line.

StandardTry it yourself first

The vertices of triangle

A BC

are

A (5, 1)

B (1, 7)

C (3, - 3)

. Find the equation of the median from

A

StandardTry it yourself first

The vertices of triangle

A BC

are

A (1, 6)

B (- 4, - 2)

C (8, 2)

. Find the equation of the altitude from

A

BC

StandardTry it yourself first

Find the perpendicular bisector of the segment

A B

, where

A (3, - 5)

and

B (- 1, 7)

StandardTry it yourself first

The vertices of a triangle are

A (0, 0)

B (8, 0)

C (2, 6)

. Using perpendicular bisectors, find the coordinates of its circumcentre.

AdvancedTry it yourself first

Find the equation of the line through the point of intersection of

x + 2 y = 9

and

3 x - y = 6

and through the point

(2, 1)

AdvancedTry it yourself first

The vertices of a triangle are

A (- 3, - 3)

B (- 1, 2)

C (2, 2)

. Find its orthocentre.

StandardTry it yourself first

Which of these points satisfies the inequality

5 x - 2 y < 10

StandardTry it yourself first

How is the region of the inequality

2 x - 3 y \geq 6

shaded in a diagram?

AdvancedTry it yourself first

Find the area of the region given by

x \geq 0

y \geq 0

and

3 x + 4 y \leq 24

StandardTry it yourself first

Find the area of the region given by

x \geq 0

y \geq 0

x \leq 4

and

x + y \leq 9

StandardTry it yourself first

How many points

(x, y)

with integer coordinates satisfy

x \geq 0

y \geq 0

and

x + 2 y \leq 8

StandardTry it yourself first

Find the area of the figure enclosed by

∣ x ∣ + ∣ y ∣ = 9

StandardTry it yourself first

Find the area enclosed by

\frac{∣ x ∣}{8} + \frac{∣ y ∣}{15} = 1

AdvancedTry it yourself first

Find the perimeter of the figure enclosed by

\frac{∣ x ∣}{8} + \frac{∣ y ∣}{15} = 1

AdvancedTry it yourself first

Find the area enclosed by

∣ x + 3∣ + ∣ y - 4∣ = 6

BasicTry it yourself first

Find the area of the region given by

∣ x ∣ \leq 5

and

∣ y ∣ \leq 9

AdvancedTry it yourself first

Find the area of the region enclosed by

x + ∣ y ∣ = 12

and the Y-axis.

StandardTry it yourself first

How many points

(x, y)

with integer coordinates satisfy

∣ x ∣ + ∣ y ∣ \leq 5

StandardTry it yourself first

A point

P (x, y)

moves so that it is always equally far from

A (4, - 1)

and

B (- 2, 5)

. Find the locus of

P

StandardTry it yourself first

Find the equation of the circle with centre

(- 1, 3)

and radius 4.

AdvancedTry it yourself first

Find the centre and radius of the circle

x^{2} + y^{2} + 6 x - 10 y - 15 = 0

AdvancedTry it yourself first

Find the equation of the circle with the points

A (- 4, 1)

and

B (2, 7)

as the ends of a diameter.

StandardTry it yourself first

Where is the point

(3, 3)

with respect to the circle

x^{2} + y^{2} + 2 y - 24 = 0

AdvancedTry it yourself first

A circle with centre

(5, - 1)

touches the line

4 x + 3 y = - 8

. Find its radius.

AdvancedTry it yourself first

Find the locus of all points at a distance of 3 units from the line

8 x + 15 y = 10

Revision

Quick questions and common mistakes

All formulas at a glance and the mistakes that cost marks.

Question	Answer
Slope of the line through two points?	$\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Slope of $a x + b y = c$ ?	$- \frac{a}{b}$
X-cut and Y-cut?	$\frac{c}{a}$ , $\frac{c}{b}$
Slope at $12 0^{\circ}$ ?	$- 3$
Triangle with the axes?	$\frac{c ^{2}}{2∣ ab ∣}$
Parallel to $a x + b y = c$ through $P$ ?	$a x + b y = a x_{1} + b y_{1}$
Perpendicular to $a x + b y = c$ through $P$ ?	$b x - a y = b x_{1} - a y_{1}$
Slopes of perpendicular lines?	$m_{1} m_{2} = - 1$
Angle between two lines?	$tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$
Distance from a point to a line?	$\frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$
Distance between parallel lines?	$\frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}}$ after equalising the coefficients
Foot and image?	$F = P - s (a, b)$ , $P^{'} = 2 F - P$
Equation of the mirror?	perpendicular bisector of $P P^{'}$
Line through the intersection of two lines?	$L_{1} + λ L_{2} = 0$
Area of $∣ x ∣ + ∣ y ∣ = k$ ?	$2 k^{2}$
Area of $\frac{∣ x ∣}{a} + \frac{∣ y ∣}{b} = 1$ ?	$2 ab$
Circle (centre, radius)?	$(x - h)^{2} + (y - k)^{2} = r^{2}$
Centre and radius of $x^{2} + y^{2} + 2 gx + 2 f y + c = 0$ ?	$(- g, - f)$ , $g^{2} + f^{2} - c$
Radius of a circle touching a line?	distance from the centre to the line

Finding the slope of a vertical line — $x = k$ has no slope; $y = k$ has slope $0$ .
Turning the slope upside down — it is $\frac{Δ y}{Δ x}$ , not the reverse.
Only reversing or only changing the sign for a perpendicular slope — do both: $- \frac{1}{m}$ .
Not making the equation “ $= 0$ ” in the distance formula — $c$ must be on the left; and not equalising the coefficients for parallel lines.
Using $\frac{c ^{2}}{2∣ ab ∣}$ for a line with $c = 0$ — a line through the origin makes no triangle.
Shading the wrong side of an inequality — do the origin test; for a line through the origin use another point.
Taking the perimeter of $∣ x ∣ + ∣ y ∣ = k$ as $4 k$ — the side is $k 2$ .
Forgetting to reverse the centre’s sign — the centre of $(x + 2)^{2}$ is $h = - 2$ ; in the general form it is $(- g, - f)$ .
Checking only one equation for concurrent lines — put the meeting point into the third line.
One solution of “ $(\dots)^{2} =$ number” — take both signs.

PracticeTen minutes a day: pick four questions from the table and say the answers without looking, then make up a new question (decide the answer first, then build the question around it — the numbers come out clean). Redo the practice questions of any method that trips you, and revise the points chapter “Coordinate Geometry” alongside.